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We would like to use the fundamental frequency of a wire to measure gravity on a new planet. We will first calibrate an instrument on Earth. We will then use the instrument on this new celestial body. a. We hang a wire having a mass of 4.30 g and length 47.9 cm from the ceiling. What mass would we hang from this wire (giving us a tension) to get a fundamental mode of vibration at 100 Hz? Assume that this wire is made of a special material that does not stretch when we hang the mass. b. We now take the same wire and the mass found in the previous section to the new planet. If we set the experiment up again and measure a new fundamental frequency of 76.6 Hz, what is g on this new planet?

Respuesta :

Answer:

Part a)

M = 8.4 kg

Part b)

[tex]g = 5.76 m/s^2[/tex]

Explanation:

As we know that wire is of mass 4.30 g and length 47.9 cm

so here we can find the mass per unit length of the wire as

[tex]\mu = \frac{m}{L}[/tex]

[tex]\mu = \frac{4.30 \times 10^{-3}}{0.479}[/tex]

[tex]\mu = 8.98 \times 10^{-3} kg/m[/tex]

now for fundamental mode of the frequency we know that

[tex]f = \frac{v}{2L}[/tex]

so we have

[tex]100 = \frac{v}{2(0.479)}[/tex]

[tex]v = 95.8 m/s[/tex]

now we have

[tex]v = \sqrt{\frac{T}{\mu}}[/tex]

so we have

[tex]95.8 = \sqrt{\frac{T}{8.98 \times 10^{-3}}}[/tex]

[tex]T = 82.4 N[/tex]

so the weight will be

[tex]Mg = 82.8 [/tex]

[tex]M = 8.4 kg[/tex]

Part b)

Now on another planet we have

[tex]f = 76.6 Hz[/tex]

so we have

[tex]v = (76.6)(2\times 0.479)[/tex]

[tex]v = 73.4 m/s[/tex]

now we have

[tex]73.4 = \sqrt{\frac{T}{8.98 \times 10^{-3}}}[/tex]

[tex]T = 48.36 N[/tex]

so the weight will be

[tex]Mg = 48.36 [/tex]

[tex]g = \frac{48.36}{8.4} = 5.76 m/s^2[/tex]

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